设总体sim N(0,1) X1,X2,···,X6是来自总体X的简单随机样本sim N(0,1) X1,X2,···,X6分别为样本均值和样本方差sim N(0,1) X1,X2,···,X6,则D(T) = ( )sim N(0,1) X1,X2,···,X6

设总体sim N(0,1) ,X1,X2,···X5为来自总体的样本,令sim N(0,1) ,X1,X2,···X5则下列说法错误的是( )sim N(0,1

22.设X1,X2,···,x6是来自总体N(0,1 )的样本,则 _(1)sim ()-|||-(A)N(0,1) (B)N(0,6) (C)x^2(1) (

总体sim N(0,1), X1,X2,X3,X4,X55,X6,X7,X8为来自总体sim N(0,1), X1,X2,X3,X4,X55,X6,X7,X8的

7.设X1,X2,···,X6是来自总体 sim N(0,1) 的简单随机样本.-|||-(1)当 k= __ _时, =dfrac (k({X)_(1)+(X

设总体X~ N(0,1),X1 , X2,…, X6为来自总体X的样本,=(({X)_(1)+(X)_(2)+(X)_(3))}^2+(({X)_(4)+(X)

4.(1)设样本x1,x2,···,X6来自总体N(0,1), =(({X)_(1)+(X)_(2)+(X)_(3))}^2+-|||-(({X)_(4)+(X

设总体 sim N(0,2) ,(X1,X2,···,X6)是来-|||-自总体X的样本,若 dfrac (a({X)_(1)+(X)_(2))}(sqrt {

4.(1)设样本x1,x2,···,X6来自总体N(0,1), =(({X)_(1)+(X)_(2)+(X)_(3))}^2+((X)_(4)+-|||-_(5

4.(1)设样本X1,X2,···X6来自总体N(0,1), =(({X)_(1)+(X)_(2)+(X)_(3))}^2+((X)_(4)+-|||-_(5)

设样本X1,X2,X3,X4,X5来自总体N(0,1),设样本X1,X2,X3,X4,X5来自总体N(0,1),设样本X1,X2,X3,X4,X5来自总体N(0